Question

39. The numerator of a fraction
is 2 less than the denominator.
If one is added to its
denominator, it becomes 1/2
find the original fraction.
*​

Answer 1

Given:

• The numerator of a fraction is 2 less than the denominator.
• If one is added to its denominator, it becomes 1/2.

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To find:

• Original Fraction?

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Solution:

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☯ Let Numerator and Denominator of a fraction be x and y respectively.

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$\underline{\bigstar\:\boldsymbol{According\:to\:the\:question\::}}$

• The numerator of a fraction is 2 less than the denominator.

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$:\implies\sf x = y - 2\qquad\qquad\bigg\lgroup\bf eq\:(1) \bigg\rgroup$

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Also,

• If one is added to its denominator, it becomes 1/2.

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$:\implies\sf \dfrac{x}{y + 1} = \dfrac{1}{2}$

$:\implies\sf \dfrac{y - 2}{y + 1} = \dfrac{1}{2}$

$:\implies\sf 2(y - 2) = y + 1$

$:\implies\sf 2y - 4= y + 1$

$:\implies\sf 2y - y = 1 + 4$

$:\implies{\underline{\boxed{\frak{\purple{y = 5}}}}}\;\bigstar$

$\dag\;{\underline{\frak{Putting\:value\:of\:y\:in\:eq\:(1),}}}$

$:\implies\sf x = 5 - 2$

$:\implies{\underline{\boxed{\frak{\purple{x = 3}}}}}\;\bigstar$

Therefore,

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• Numerator of the fraction, x = 3
• Denominator of the fraction, y = 5

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$\therefore\:{\underline{\sf{Hence,\:the\;original\: fraction\:is\: \bf{ \dfrac{3}{5}}.}}}$

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$\qquad\qquad\boxed{\underline{\underline{\pink{\bigstar \: \bf\: Verification\:\bigstar}}}}$

• If one is added to its denominator, the fraction becomes 1/2.

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$:\implies\sf \dfrac{3}{5 + 1} = \dfrac{1}{2}$

$:\implies\sf \cancel{\dfrac{3}{6}} = \dfrac{1}{2}$

$:\implies\sf \dfrac{1}{2} = \dfrac{1}{2}$

$\qquad\qquad\qquad\bf{\dag}\:{\underline{\underline{\sf{\purple{Hence\: Verfied!}}}}}$

Answer 2

Answer:

Given :-

• The numerator of a fraction is 2 less than the denominator. If one is added to its denominator its become ½.

To Find :-

• What is the original fraction.

Solution :-

Let, the numerator be x

And, the denominator will be x + 2

Hence the original fraction will be $\sf\dfrac{x}{x + 2}$

According to the question,

⇒ $\sf\dfrac{x}{x + 2 + 1} =\: \dfrac{1}{2}$

⇒ $\sf\dfrac{x}{x + 3} =\: \dfrac{1}{2}$

By doing cross multiplication we get,

⇒ 2(x) = x + 3

⇒ 2x = x + 3

⇒ 2x - x = 3

➠ x = 3

Hence, the required original fraction will be,

↦ $\sf\dfrac{x}{x + 2}$

↦ $\sf\dfrac{3}{3 + 2}$

➦ $\sf\dfrac{3}{5}$

$\therefore$ The original fraction is $\sf\boxed{\bold{\large{\dfrac{3}{5}}}}$ .

Let's Verify :-

⇒ $\sf\dfrac{x}{x + 2 + 1} =\: \dfrac{1}{2}$

Put x = 3 we get,

⇒ $\sf\dfrac{3}{3 + 2 + 1} =\: \dfrac{1}{2}$

⇒ $\sf\dfrac{3}{6} =\: \dfrac{1}{2}$

⇒ $\sf\dfrac{\cancel{3}}{\cancel{6}} =\: \dfrac{1}{2}$

⇒ $\sf\dfrac{1}{2} =\: \dfrac{1}{2}$

➦ LHS= RHS

Hence, Verified ✔